Question

Let $f\left(x\right)$ be a periodic function with period $3$ and $f(-\frac{2}{3})=7$ and $g\left(x\right)={\int }_0^{x}f(t+n)dt$, where $n=3K,K\in N$. Then, $g^{\prime }\left(\frac{7}{8}\right)$ is equal to

• A. $-\frac{2}{3}$
• B. $7$
• C. $-7$
• D. $\frac{7}{3}$

Answer 1

The correct option is B $7$

$g^{\prime }\left(x\right)=f(x+n)=f(x+(n-3)+3)$
$f(x+n-3)$
$=f(x+3K-3)$
$=f(x+3(K-1\left)\right)$

Since, $K\in N$,
$=f(x+3(K-1\left)\right)=f\left(x\right)$

$\Rightarrow g^{\prime }\left(x\right)=f\left(x\right)$

$\Rightarrow g^{\prime }\left(\frac{7}{3}\right)=f\left(\frac{7}{3}\right)=f(-\frac{2}{3}+3)=f(-\frac{2}{3})=7$
Hence, (b) is the correct answer.